A few families took a trip to an amusement park together. Tickets cost $$8.00$ each for adults and $$3.50$ each for kids, and the group paid $$45.00$ in total. There were $3$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Answer: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${8x+3.5y = 45}$ ${x = y-3}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-3}$ for $x$ in the first equation. ${8}{(y-3)}{+ 3.5y = 45}$ Simplify and solve for $y$ $ 8y-24 + 3.5y = 45 $ $ 11.5y-24 = 45 $ $ 11.5y = 69 $ $ y = \dfrac{69}{11.5} $ ${y = 6}$ Now that you know ${y = 6}$ , plug it back into ${x = y-3}$ to find $x$ ${x = }{(6)}{ - 3}$ ${x = 3}$ You can also plug ${y = 6}$ into ${8x+3.5y = 45}$ and get the same answer for $x$ ${8x + 3.5}{(6)}{= 45}$ ${x = 3}$ There were $3$ adults and $6$ kids.